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If a speed man is rotated at the end of a long beam of length 5m and the acceleration of 9g, then the number of revolution performed will be (`g=10 m//s^(2))`A. `(3)/(sqrt(2)pi) rps`B. `(3pi)/(sqrt2) rps`C. `(2sqrt5)/(3) rps`D. `(sqrt3)/(3 pi ) rps` |
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Answer» Correct Answer - A Centripetal acceleration `(v^(2))/(r)= 9g ` `r omega ^(2)=9g` `omega^(2)=(9g)/(r)` `omega=sqrt((9g)/(r))` `2pi n = sqrt((9g)/(r))` `n=(1)/(2pi)sqrt((9g)/(r))`. |
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